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Steph_Rawr352
*~*~*~*HELP PLEASE*~*~*~* Factor 400x^2 + 160x + 16 completely.
Oh.. and I have to factor it using a special product or something?
a=400, b=160, c=16 ac=6400 We need to find factors of 6400 that add up to 160 do start the problem.
Ok. Now we will form this square:\[\left[\begin{matrix}400x^{2} & 80x \\ 80x & 16\end{matrix}\right]\]Notice that ax^2 is the first element of the first row, c is the second element of the second row, and the two factors times x are the other two entries (it doesn't matter which spot you put the factors in). Now, the next step is to find the greatest common factors of each row and each column. For example, the row with 80x and 16 has 16 as its greatest common factor. (In your work write this in front of the that row. I would show you exactly what I mean, but I don't think the formatting here will let me do it.) Then do that with the rest of the rows and columns. Finally, take those factors that you found and they will be the factored solution. The factors for the columns will be one set of parenthesis, and the factors for the rows will be the other set.
400x^2 + 160x + 16 = (20x)^2 +160x+4^2 = (20x)^2 +2(20x)(4)+4^2 can you do it now??
hint: a^2+2ab+b^2 = (a+b)^2 = (a+b)(a+b)
The only other thing to expand it to a more general case (where you would say have negative factors or a is negative) is that the factor you write in front of the row or above the column has the same sign as the closest element to it. So for example if a was negative, the factor for that row would be negative, regardless of whether or not the other entry in that row was negative or positive. The same holds true for that column.
@Aylin I don't understand the square thing with the ax^2 and c? and @rizwan_uet where would you put the factors that add up to 160?
That's just a box you put the entries into to help with factoring.
When you have your final answer you can post it here and either I or rizwan should be able to tell you if it is correct or give some more help if needed.
ohk i will(: Thanks guys I wish I could pick 2 best answers
I got (20x + 4)^2
That does work, but it is not the simplest form. 20x and 4 both share a factor of 4. Do you know how to make it simpler?
so it would be 5x + 1?
Sort of. \[(20x+4)^{2}=(4(5x+1))^{2}=16(5x+1)^{2}\]
oh ohk thank you!!
you have this ok (20x)^2 +160x+4^2 now from the factors (20x)^2 and 4^2 i can write (20x)^2 +2(20x)(4)+4^2 because i know the formula a^2+2ab+b^2 = (a+b)^2 thus (20x)^2 +2(20x)(4)+4^2 = (20x+4)^2
now (20x+4)^2 = (20x+4)(20x+4) thus the required factors are (20x+4)(20x+4)
oh I understand now thank you so much!